How do you prove $\frac{sin x + cos x}{sec x + csc x} = \frac{sin x}{sec x}$ $(sinx+cosx)/(secx+cscx)=sinx/secx$ ?

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Answer 1 · 2016-08-25T14:59:53

See below.

Apply the following identities:

• $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$

• $csc θ = \frac{1}{sin θ}$ $csctheta = 1/sintheta$

Now, simplify both sides using the given identities:

$\frac{sin x + cos x}{\frac{1}{cos x} + \frac{1}{sin x}} = \frac{sin x}{\frac{1}{cos x}}$ $(sinx + cosx)/(1/cosx + 1/sinx) = sinx/(1/cosx)$

$\frac{sin x + cos x}{\frac{sin x + cos x}{sin x cos x}} = sin x cos x$ $(sinx + cosx)/((sinx + cosx)/(sinxcosx)) = sinxcosx$

$sin x + cos x \times \frac{sin x cos x}{sin x + cos x} = sin x cos x$ $sinx + cosx xx (sinxcosx)/(sinx + cosx) = sinxcosx$

$cancel(sinx + cosx) xx (sinxcosx)/(cancel(sinx+cosx)) = sinxcosx$

$sinxcosx = sinxcosx$

Identity proved!!

Hopefully this helps!

How do you prove (sinx+cosx)/(secx+cscx)=sinx/secxsinx+cosxsecx+cscx=sinxsecx(sinx+cosx)/(secx+cscx)=sinx/secx?